STOICHIOMETRY

     1.      Understanding Stoichiometry

Stoichiometry comes from two Greek syllables Stoicheion meaning "element" and Metron which means "measurement". Stoichiometry is a subject in chemistry involving the linkage of reactants and products in a chemical reaction to determine the quantity of each reacting agent.

2.  BASIC LAW OF CHEMICAL SCIENCE

A. The Law of Conservation of Mass (Lavoisier's Law)

"The mass of substances before and after the reaction is the same".

Example:

S + O 2 → SO 2

2 gr 32 gr 64 gr

B. Fixed Comparative Law (Proust Law)

"The ratio of the elemental masses in each compound is fixed"

Example:

H 2 O → mass H: mass O = 2: 16 = 1: 8

C. The Law of Multiple Comparisons (Dalton's Law)

"If two elements can form two or more compounds, and the mass of one element is equal, the ratio of the mass of the second element is proportional to the simple and integer".

Example:

- Element N and O can form NO and NO 2 compounds

- In the NO compound, mass N = mass O = 14: 16

- In the compound NO 2, mass N = mass O = 14: 32

- The ratio of N masses to NO and NO 2 is the same

O = 16: 32 = 1: 2 mass ratio

D. Ideal Gas Law

For an ideal gas or a gas that is considered to be valid applies the formula:

PV = n RT

Information:P = pressure (atmosphere)V = volume (liters)N = mol = gram / MrR = gas constant (lt.atm / mol.K)T = temperature (Kelvin)

3.  Mol (n)  

The atom is the smallest part that constitutes an element, while the molecule is the smallest part that constitutes a compound. The next atom and molecule are called elementary particles. The international unit for atoms and molecules is mol. One mole of the substance is the number of substances containing elementary particles of Avogadro (L), which is 6.02 x 1023. The number of moles is expressed by the symbol n. 1 mol element = 6.02 x 1023 atoms of that element 1 mol of compound = 6.02 x 1023 molecule of the compound

Thus n = number of particles / (6.02 X 1023)

 as an example : 1 mol H2O = 1 X (6.02 X 1023) Means in 1 mole of H2O there is 6,02 X 1023

4.  Relative atomic mass and molar mass

A. Relative Atom Mass (Ar), 
In chemical calculations not used absolute mass but used relative atomic mass (Ar). The relative atomic mass (Ar) is th ratio of the average mass of one atom of an element to 1/12 atomic mass of 12C or 1 sma (atomic mass unit) = 1.66 x 10-24 grams.

 Example: Ar H = 1.0080 sma rounded 1

Ar C = 12.01 sma rounded 12

Ar N = 14,0067 sma rounded

14 Ar O = 15,9950 sma rounded 16

The list of relative atomic masses (Ar) can be seen in the periodic table

B. Relative Molecular Mass (Mr)
The relative molecular mass (Mr) is a number representing the mass ratio of one molecule of a compound to 1/12 of the atomic mass of 12C. The molecular mass of a realtif (Mr) equals the sum of the relative atomic mass (Ar) of all the constituent atoms.

Example: Mr. H2O = (2 x Ar H) + (1 X Ar O) = (2 x 1) + (1 x 16) = 18 Mr. CO (NH2) 2 = (1 x Ar C) + (1 x Ar O) + (2 x Ar N) + (4 x Ar H) = (1 x 12) + (1 X 16) + (2 X 14) + (4 x 1) = 60

C. Molar Mass

The mass of one mol of element or the mass of one mole of a compound is called the molar mass. 

The mass of one mole of element is equal to the relative atomic mass (Ar) of the atom in grams, 

whereas the mass of one mole of the compound is equal to the relative molecular mass (M) of the 

compound in grams

So n (mol) = mass (gram) / Ar or Mr (gram / mol)

Another example, on the periodic table, we can see that the mass of one copper atom is 63,55 sma 

and the mass of one sulfur atom is 32.07 sma. Meanwhile, the mass of one oxygen atom is 16.00 

sma, while the mass of one hydrogen atom is 1.008 sma. Thus, the mass of one molecule of 

CuSO4.5H2O is as follows:

Mr. CuSO4.5H2O = 1 x Ar Cu + 1 x Ar S + 4 x Ar O + 5 x Mr H2O

= 1 x Ar Cu + 1 x Ar S + 4 x Ar O + 5 x (2 x Ar H + 1 X Ar O)

= 1 x 63,55 + 1 x 32,07 + 4 x 16,00 + 5 x (2 x 1,008 + 1 x 16,00)

= 249,700 sma

D. Molar Volumes  

The molar volume is the volume of one mole of gas. One mole of gas contains 6.02 x 1023 molecules. That is, every gas that has the same number of molecules, the number of moles is the same. In accordance with Avogadro's law, at the same temperature (T) and pressure (P), all gases of the same volume (V) contain the same number of moles (n).

Application of Avogadro's Law in Various Circumstances

A.    Circumstance in Standard Temperature and Pressure (STP)  

Based on the Avogadro hypothesis and the ideal gas equation, the volume of 1 mol of each gas in the standard state (STP), ie at P = 1 atm and T = 0°C = 273 K is 22.4 liters. So n (mol) = volume / 22.4 (liter / mol)   As an example : Determine the volume of 5 moles of carbon dioxide (CO2) gas measured on the STP. Means Volume = 5 X 22.4 Liter = 112

B. Circumstances in Temperature and Non-Standard Pressure

In non-standard circumstances, the molar volume is calculated by the ideal gas equation PV = nRT 

(T in Kelvin)

C. Circumstances on Other Known Temperature and Gas Pressure

At the same temperature and pressure, the same volumes of gas have the same number of moles, 

so that the volume ratio at the same temperature and temperature will be equal to the mole ratio.

Thus, V1 / V2 = n1 / n

V1 = volume of gas 1

V2 = gas volume 2

N1 = number of moles of gas 1

N2 = the number of moles of gas 2

5. STOIKIOMETR

I Stoichiometry comes from the Greek word stoicheion meaning element and metron which means measure. Stoichiometry discusses the relation of mass between elements in a compound (stoichiometric compound) and interactivity in a reaction (reaction stoichiometry

1.      Composition of Substances

The mass ratio of the constituent elements of a compound is always fixed, so it can be calculated the percentage of the elements in a compound. The percentage of elements in a compound is based on the ratio of the relative atomic mass amount (Ar) of a given element to the relative molecular mass (MR) of the compound. Problems example § Determine the percentage of elements C, H, and O in glucose (C6H12O6) (Ar C = 12, H = 1, and O = 16).

2. Determination of Empirical Formulas and Molecular Formulas

The empirical formula is the simplest comparison formula of the atoms of various elements in the 

compound. Steps to determine the empirical formula of a compound:

A. Determine the mass of each element in the compound

B. Dividing the mass of each element with its relative atomic mass to obtain the mole ratio of each 

element

C. Converts the mole ratio to simple numbers

The molecular formula describes the number of atoms of each element that make up the molecule 

of compound. The molecular formula is a multiple of the empirical formula. If the empirical 

formula and molecular mass are relatively known, then the molecular formula can be determined.  

Problems example

  § 1.5 grams of a compound containing 0.3 grams of hydrogen and 1.2 grams of carbon. If the relative 

molecular mass of the compound is 30, find the empirical formula and the molecular formula.

Comparison of the number of moles C: the number of moles H = 0.1: 0.3 = 1: 3 so that the empirical 

formula of the compound is CH3 or CH3

The relative molecular mass (Mr) of the compound = 30

(CH3) n = 30

(Ar C + 3 Ar H) n = 30

(12 + 3) n = 30

15 n = 30 so n = 2

Thus, the molecular formula of the compound is (CH3) n or C2H6

3. Chemical calculations

In accordance with the Law of Volume Comparison (Gay Lussac), the ratio of the volume of gases 

corresponds to the respective coefficients of the gas. Since the volume ratio is in accordance with 

the mole ratio, it can be said that the ratio of the number of moles of the substance corresponds to 

the ratio of the coefficients of each substance.

The reaction coefficient is the ratio of the number of particles of the substance involved in the reaction. 

Since 1 mole of each substance contains the same number of particles, the ratio of the number of 

particles is equal to the ratio of the number of moles. Thus, the reaction coefficient is the ratio of the 

number of moles of substances involved in the reaction.

Consider the following reactions:

N2 (g) + 3 H2 (g) → 2 NH3 (g)

The reaction coefficient states that 1 molecule of N2 reacts with 3 molecules of H2 to form 2 molecules

 of NH3 or 1 mole of N2 reacts with 3 moles of H2 yielding 2 moles of NH 3. In that sense, the amount 

of substances required or produced in a chemical reaction can be calculated using equivalent reaction 

equations. When the number of moles of one of the reacting substances is known, then the number of 

moles of the other substance in the reaction can be determined by using the ratio of the reaction coeffi

cient.

To complete the chemical calculations,

The steps are as follows:

A) Write down the equation of the reaction and equalize its coefficients.

B) Converts a unit of substance known to mole.

C) Find the mole of the substance in question by comparing the coefficients.

D) Converting mol units into other desired units.

4. Limiting Reagents

In a chemical reaction, the mole ratio of the added reagents is not always the same as the ratio of the 

reaction coefficient. This causes a reagent to be reacted first. Such reagents are called limiting reagents.

Problems example

One mole of sodium hydroxide (NaOH) solution is reacted with one mole of sulfuric acid (H 2 SO 4) 

solution according to the reaction:

2 NaOH (aq) + H2SO4 (aq) ®Na2SO4 (aq) + 2 H2O (l)

Specify:

A. Limiting Reagents

B. The remaining reagents

C. Mol Na2SO4 and mole of H2O produced

Then

NaOH: H2SO4 = 2: 1, ja H2 in every two parts NaOH will react with one part H2SO4

Since the amount of NaOH and H2SO4 does not meet the comparison, then there are reactants that 

have reacted and there are reagents remaining.

NaOH: H2SO4 = 2: 1, so every two moles of NaOH requires one mole of H2SO4 or every mole of 

NaOH requires 0.5 mole of H2SO4